Question: Find the value of $c$ so that the polynomial $p(x)$ is divisible by $(x-4)$. $p(x) = cx^3-15x -68$ $c=$
Explanation: The following statements are equivalent: $(x-4)$ is a factor of $p(x)$ $p(x)$ is divisible by $(x-4)$ The remainder of $\dfrac{p(x)}{x-4}$ is $0$ We can use the polynomial remainder theorem to solve this problem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x-a$ is $p(a)$. According to the theorem, the remainder when $p(x)$ is divided by $(x-{4})$ is equal to $p({4})$. We want this remainder to be equal to $0$. So let's set $p({4})=0$ and solve this equation to find $c$. Let's plug ${x=4}$ in $p( x) = c x^3-15 x-68$ and set that equal to $0$. $\begin{aligned} c({4})^3-15({4}) -68&=0 \\\\ 64c-60-68&=0 \\\\ 64c - 128&=0 \\\\ 64c&=128 \\\\ c&=2 \end{aligned}$ To conclude, $c=2$